## A Curious Geometric Limit

Inspired by Alexander Bogomolny (@CutTheKnotMath) and his curious geometric limit, I offer this trigonometric approach.

Consider the following diagram in which we have circle X of radius 3 centered at (3,0) and circle O of radius k centered at the origin.  Call the intersection of the these circles in the first quadrant B.  Let A be the intersection of circle O with the y-axis, and extend line AB until it intersects the x-axis at E.  Our goal is to show that $\lim_{kto}OE=12$.

Consider the diagram below.

Let $\angle{OXB} = \theta$.  Some simple angle chasing, using properties of isosceles triangles and right triangles, gives us that $\angle{OEA} = \frac{\theta}{4}$.

Since AO = k by definition, we have $tan(\frac{\theta}{4}) = \frac{k}{OE}$, and so OE = $\frac{k}{tan(\frac{\theta}{4}) }$.

Since OB = k is the base of isosceles triangle OXB, we can drop an altitude from X to OB and find that $sin(\frac{\theta}{2}) = \frac{\frac{k}{2}}{BX} = \frac{\frac{k}{2}}{3}$.    This gives us $k = 6sin(\frac{\theta}{2})$.

So OE =  $\frac{k}{tan(\frac{\theta}{4}) }$ =  $\frac{6sin(\frac{\theta}{2})}{tan(\frac{\theta}{4}) }$.  By the double angle formula for sine, we have

$sin(\frac{\theta}{2}) = 2sin(\frac{\theta}{4})cos(\frac{\theta}{4})$,

Making this substitution, and by writing tangent as the quotient of sine and cosine, we get

OE = $\frac{6sin(\frac{\theta}{2})}{tan(\frac{\theta}{4}) } = \frac{12sin(\frac{\theta}{4}) cos((\frac{\theta}{4}))}{\frac {sin(\frac{\theta}{4})}{cos(\frac{\theta}{4})} } = 12 {(cos(\frac{\theta}{4}))}^2$.

Finally, since $\theta\to 0$ as $k \to 0$, we have

$\lim_{k\to 0}OE = \lim_{\theta \to 0}12 {(cos(\frac{\theta}{4}))}^2 = 12\lim_{\theta \to 0} {(cos(\frac{\theta}{4}))}^2=12$.